Q7 of 230 Page 29

Find a normal unit vector to the plane x + 2y + 3z–6 = 0.

The given equation of the plane is x + 2y + 3z–6 = 0

x + 2y + 3z = 6



or,


where ……(i)


Now,


Unit vector to the plane,


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5

Write the normal form of the equation of the plane

2x–3y + 6z + 14 = 0.

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Find the equation of a plane which is at a distance of 3√3 units from the origin and the normal to which is equally inclined with the coordinate axes.

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